What is the sum of all integer values $n$ for which $\binom{26}{13}+\binom{26}{n}=\binom{27}{14}$?
Solution: From Pascal's identity $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}$.

Therefore, we have $\binom{26}{13}+\binom{26}{14}=\binom{27}{14}$, so $n=14$.

We know that $\binom{27}{14}=\binom{27}{27-14}=\binom{27}{13}$.

We use Pascal's identity again to get $\binom{26}{13}+\binom{26}{12}=\binom{27}{13}$, so $n=12$.

There are two values for $n$, $12$ and $14$, so the sum is $12+14=\boxed{26}$.